略

设 $A(x_{1},y_{1})$，$B(x_{2},y_{2})$，$C(x_{3},y_{3})$，$D(x_{4},y_{4})$，$\overrightarrow{AP}=\lambda \overrightarrow{PC}$，则有$$x_{0}-x_{1}=\lambda (x_{3}-x_{0}),y_{0}-y_{1}=\lambda(y_{3}-y_{0}),$$所以\[x_{3}=\dfrac{(1+\lambda )x_{0}-x_{1}}{\lambda}, y=\dfrac{(1+\lambda)y_{0}-y_{1}}{\lambda}.\]因为点 $C$ 在椭圆上，所以$$\dfrac{x_{3}^{2}}{4}+y_{3}^{2}=1,$$即\[\dfrac{[(1+\lambda )x_{0}-x_{1}]^{2}}{4\lambda^{2}}+\dfrac{[(1+\lambda )y_{0}-y_{1}]^{2}}{\lambda ^{2}}=1,\]整理得\[(1+\lambda)^{2}\left(\dfrac{x_{0}^{2}}{4}+y_{0}^{2}\right)-\dfrac{1}{2}(1+\lambda)(x_{0}x_{1}+4y_{0}y_{1})+\left(\dfrac{x_{1}^{2}}{4}+y_{1}^{2}\right)=\lambda^{2}.\]又点 $A$ 在椭圆上，所以$$\dfrac{x_{1}^{2}}{4}+y_{1}^{2}=1,$$从而可得\[(1+\lambda)^{2}\left(\dfrac{x_{0}^{2}}{4}+y_{0}^{2}\right)-\dfrac{1}{2}(1+\lambda)(x_{0}x_{1}+4y_{0}y_{1})=\lambda^{2}-1\cdots\cdots \text{ ① }\]又因为 $AB\parallel CD$，故有 $\overrightarrow{BP}=\lambda \overrightarrow{PD}$，同理可得\[(1+\lambda)^{2}\left(\dfrac{x_{0}^{2}}{4}+y_{0}^{2}\right)-\dfrac{1}{2}(1+\lambda)(x_{0}x_{2}+4y_{0}y_{2})=\lambda^{2}-1\cdots\cdots \text{ ② }\]② $-$ ① 得，\[x_{0}(x_{1}-x_{2})+4y_{0}(y_{1}-y_{2})=0.\]因为 $x_{0}\ne 0$，$y_{0}\ne 0$，易知 $AB$ 不与坐标轴平行，所以直线 $AB$ 的斜率$$k=\dfrac{y_{1}-y_{2}}{x_{1}-x_{2}}=-\dfrac{x_{0}}{4y_{0}}$$为定值．

略

直线 $EF$ 的方程为$$y=-\dfrac{x_{0}}{4y_{0}}(x-x_{0})+y_{0},$$代入椭圆方程得\[\dfrac{x^{2}}{4}+\left[-\dfrac{x_{0}}{4y_{0}}(x-x_{0})+y_{0}\right]^{2}=1,\]整理得\[\dfrac{x_{0}^{2}+4y_{0}^{2}}{16y_{0}^{2}}\cdot x^{2}-\dfrac{x_{0}(x_{0}^{2}+4y_{0}^{2})}{8y_{0}^{2}}\cdot x+\dfrac{x_{0}^{4}}{16y_{0}^{2}}+\dfrac{x_{0}^{2}}{2}+y_{0}^{2}-1=0,\]所以\[x_{E}+x_{F}=-\dfrac{-\dfrac{x_{0}(x_{0}^{2}+4y_{0}^{2})}{8y_{0}^{2}}}{\dfrac{x_{0}^{2}+4y_{0}^{2}}{16y_{0}^{2}}}=2x_{0}.\]因此点 $P$ 是 $EF$ 的中点，即点 $P$ 平分线段 $EF$．